Question

A mould cavity of $1200\text{ cm}^3$ volume has to be filled through a sprue of $10\text{ cm}$ length feeding a horizontal runner. Cross-sectional area at the base of the sprue is $2\text{ cm}^2$. Consider acceleration due to gravity as $9.81\text{ m/s}^2$. Neglecting frictional losses due to molten metal flow, the time taken to fill the mould cavity is}

• A. 4.28 s
• B. 5.26 s
• C. 7.23 s
• D. 5.36 s

Hint:

Always keep your units consistent.
Converting gravity $g = 9.81\text{ m/s}^2$ directly to $981\text{ cm/s}^2$ avoids the need to convert volume from $\text{cm}^3$ to $\text{m}^3$, saving computation steps.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is a casting design problem where we need to find the total time required to completely fill a specified mold cavity volume through a gravity-fed gating system.

Step 2: Key Formula or Approach:
The velocity of molten metal at the base of the sprue of height $h$ is given by Torricelli's theorem:
\[ v = \sqrt{2 \cdot g \cdot h} \]
The volumetric flow rate ($Q$) through the gate is:
\[ Q = A \cdot v \]
where $A$ is the cross-sectional area at the base of the sprue.
The time taken to fill the mold cavity of volume $V$ is:
\[ t = \frac{V}{Q} \]

Step 3: Detailed Explanation:

• Convert all given parameters to consistent units (centimeters and seconds):
- Cavity volume, $V = 1200 \text{ cm}^3$
- Height of sprue, $h = 10 \text{ cm} = 0.1 \text{ m}$
- Cross-sectional area, $A = 2 \text{ cm}^2$
- Gravitational acceleration, $g = 9.81 \text{ m/s}^2 = 981 \text{ cm/s}^2$

• Calculate the velocity of the molten metal at the base of the sprue:
\[ v = \sqrt{2 \times 981 \text{ cm/s}^2 \times 10 \text{ cm}} = \sqrt{19620 \text{ cm}^2\text{/s}^2} \approx 140.07 \text{ cm/s} \]

• Compute the volumetric flow rate ($Q$):
\[ Q = A \cdot v = 2 \text{ cm}^2 \times 140.07 \text{ cm/s} = 280.14 \text{ cm}^3\text{/s} \]

• Determine the total filling time ($t$):
\[ t = \frac{V}{Q} = \frac{1200 \text{ cm}^3}{280.14 \text{ cm}^3\text{/s}} \approx 4.283 \text{ s} \]

Step 4: Final Answer:
The time taken to fill the mould cavity is approximately $4.28 \text{ s}$.